The equation of a circle $C$ is $x^2+y^2+12x+6y+29 = 0$. What is its center $(h, k)$ and its radius $r$ ?
To find the equation in standard form, complete the square. $(x^2+12x) + (y^2+6y) = -29$ $(x^2+12x+36) + (y^2+6y+9) = -29 + 36 + 9$ $(x+6)^{2} + (y+3)^{2} = 16 = 4^2$ Thus, $(h, k) = (-6, -3)$ and $r = 4$.